Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{5k^3 - 25k^2 - 250k}{-2k^3 + 2k^2 + 60k} \div \dfrac{k - 4}{3k - 18} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{5k^3 - 25k^2 - 250k}{-2k^3 + 2k^2 + 60k} \times \dfrac{3k - 18}{k - 4} $ First factor out any common factors. $a = \dfrac{5k(k^2 - 5k - 50)}{-2k(k^2 - k - 30)} \times \dfrac{3(k - 6)}{k - 4} $ Then factor the quadratic expressions. $a = \dfrac {5k(k + 5)(k - 10)} {-2k(k + 5)(k - 6)} \times \dfrac {3(k - 6)} {k - 4} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac { 5k(k + 5)(k - 10) \times 3(k - 6)} { -2k(k + 5)(k - 6) \times (k - 4)} $ $a = \dfrac {15k(k + 5)(k - 10)(k - 6)} {-2k(k + 5)(k - 6)(k - 4)} $ Notice that $(k + 5)$ and $(k - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {15k\cancel{(k + 5)}(k - 10)(k - 6)} {-2k\cancel{(k + 5)}(k - 6)(k - 4)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $a = \dfrac {15k\cancel{(k + 5)}(k - 10)\cancel{(k - 6)}} {-2k\cancel{(k + 5)}\cancel{(k - 6)}(k - 4)} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $a = \dfrac {15k(k - 10)} {-2k(k - 4)} $ $ a = \dfrac{-15(k - 10)}{2(k - 4)}; k \neq -5; k \neq 6 $